﻿Nemytskii Operator in the Space of Set-Valued Functions of Bounded <i>φ</i>-Variation

Vol.3 No.6(2013), Article ID:36373,13 pages DOI:10.4236/apm.2013.36072

Nemytskii Operator in the Space of Set-Valued Functions of Bounded -Variation

Departamento de Fsica y Matemática, Universidad de Los Andes, Trujillo-Venezuela, RBV

Received June 13, 2013; revised July 23, 2013, accepted August 19, 2013

Keywords: Bounded Variation; Function of Bounded Variation in the Sense of Riesz; Variation Space; Weight Function; Banach Space; Algebra Space

ABSTRACT

In this paper we consider the Nemytskii operator, i.e., the composition operator defined by, where is a given set-valued function. It is shown that if the operator maps the space of functions bounded -variation in the sense of Riesz with respect to the weight function into the space of set-valued functions of bounded -variation in the sense of Riesz with respect to the weight, if it is globally Lipschitzian, then it has to be of the form, where is a linear continuous set-valued function and is a set-valued function of bounded -variation in the sense of Riesz with respect to the weight.

1. Introduction

In [1], it was proved that every globally Lipschitz Nemytskii operator

mapping the space into itself admits the following representation:

where is a linear continuous set-valued function and is a set-valued function belonging to the space. The first such theorem for singlevalued functions was proved in [2] on the space of Lipschitz functions. A similar characterization of the Nemytskii operator has also been obtained in [3] on the space of set-valued functions of bounded variation in the classical Jordan sense. For single-valued functions it was proved in [4]. In [5,6], an analogous theorem in the space of set-valued functions of bounded -variation in the sense of Riesz was obtained. Also, they proved a similar result in the case in which that the Nemytskii operator N maps the space of functions of bounded -variation in the sense of Riesz into the space of set-valued functions of bounded -variation in the sense of Riesz, where, and is globally Lipschitz. In [7], they showed a similar result in the case where the Nemytskii operator maps the space of setvalued functions of bounded -variation in the sense of Riesz into the space of set-valued functions of bounded -variation in the sense of Riesz and is globally Lipschitz.

While in [8], we generalize article [6] by introducing a weight function. Now, we intend to generalize [7] in a similar form we did in [8], i.e., the propose of this paper is proving an analogous result in which the Nemytskii operator maps the space of setvalued functions of bounded -variation in the sense of Riesz with a weight into the space of set-valued functions of bounded -variation in the sense of Riesz with a weight and is globally Lipschitz.

2. Preliminary Results

In this section, we introduce some definitions and recall known results concerning the Riesz -variation.

Definition 2.1 By a -function we mean any nondecreasing continuous function such that if and only if, and as.

Let be the set of all convex continuous functions that satisfy Definition 2.1.

Definition 2.2 Let be a normed space and be a -function. Given be an arbitrary (i.e., closed, half-closed, open, bounded or unbounded) fixed interval and a fixed continuous strictly increasing function called a it is weight. If, we define the (total) generalized -variation of the function with respect to the weight function in two steps as follows (cf. [9]). If is a closed interval and is a partition of the interval I (i.e.,), we set

Denote by the set of all partitions of, we set

If is any interval in, we put

The set of all functions of bounded generalized - variation with weight will be denoted by

.

If, and, , , the -variation, also written as, is the classical -variation of in the sense of Riesz [10], showing that if and only if (i.e., is absolutely continuous) and its almost everywhere derivative is Lebesgue -summable on. Recall that, as it is well known, the space with I, and as above and endowed with the norm

is a Banach algebra for all.

Riesz’s criterion was extended by Medvedev [11]: if, then if and only if

and. Functions of bounded generalized -variation with and (also called functions of bounded Riesz-Orlicz -variation) were studied by Cybertowicz and Matuszewska [12]. They showed that if, then

and that the space

is a semi-normed linear space with the LuxemburgNakano (cf. [13,14]) seminorm given by

.

Later, Maligranda and Orlicz [15] proved that the space equipped with the norm

is a Banach algebra.

3. Generalization of Medvedev Lemma

We need the following definition:

Definition 3.1 Let be a -function. We say satisfies condition if

(1)

For φ convex, (1) is just. Clearlyfor the space coincides with the classical space of functions of bounded variation. In the particular case when and

, we have the space of functions of bounded Riesz -variation. Let

be a measure space with the Lebesgue-Stieltjes measure defined in -algebra and

Moreover, let be a function strictly increasing and continuous in. We say that has - measure 0, if given there is a countable cover by open intervals of, such that

.

Since is strictly increasing, the concept of “measure” coincides with the concept of “measure 0” of Lebesgue. [cf. [16], 25].

Definition 3.2 (Jef) A function is said to be absolutely continuous with respect to, if for every, there exists such that

for every finite number of nonoverlapping intervals, with and

.

The space of all absolutely continuous functions, with respect to a function strictly increasing, is denoted by. Also the following characterization of [17,18] is well-known:

Lemma 3.3 Let. Then exists and is finite in, except on a set of -measure.

Lemma 3.4 Let. Then is integrable in the sense Lebesgue-Stieltjes and

Lemma 3.5 Let such that satisfies the

condition. If, then is -absolutely continuous in, i.e.,

Also the following is a generalization of Medvedev Lemma [11]:

Theorem 3.6 (Generalization a Medvedev Lemma) Let such that satisfies the condition,. Then 1) If is -absolutely continuous on and

then

and

.

2) If (i.e.,), then is -absolutely continuous on and

.

Proof.) Since is absolutely continuous, there exists a.e. in by Lemma 3.3. Let,

by Lemma 3.4 and is strictly increasing

using the generalized Jenssen’s inequality

Let be any partition of interval; then

and we have

.

Thus.

) Let. Then is -absolutely continuous on by Lemma 3.5 and exist a.e. on.

For every, we consider

a partition of the interval define by

,.

Let be a sequence of step functions, defined by

converge to a.e. on. It is sufficient to prove in those points where is - differentiable and different from, for, i.e., in

For, and each, there exists such that, so

Therefore, is a convex combination of points

Now if, then and and since is -differentiable for, the expressions

tend to which is -differentiable from in. So results

Since is continuous, we have

Using the Fatou’s Lemma and definition of sequence, results that

By definition from

which is what we wished to demonstrate.

Corollary 3.7 Let such that satisfies the condition, then if and only if is -absolutely continuous on and

.

Also

Corollary 3.8 Let such that satisfies the condition. If, then is -absolutely continuous on and

4. Set-Valued Function

Let be the family of all non-empty convex compact subsets of and be the Hausdorff metric in, i.e.,

where, or equivalently,

where

(2)

Definition 4.1 Let, a fixed continuous strictly increasing function and. We say that has bounded -variation in the sense of Riesz if

(3)

where the supremum is taken over all partitions of.

Definition 4.2 Denote by

(4)

and

(5)

both equipped with the metric

(6)

where

Now, let, be two normed spaces and be a convex cone in. Given a set-valued function we consider the Nemytskii operator generated by, that is the composition operator defined by:

We denote by the space of all setvalued function, i.e., additive and positively homogeneous, we say that is linear if.

In the proof of the main results of this paper, we will use some facts which we list here as lemmas.

Lemma 4.3 ([19]) Let be a normed space and let be subsets of. If are convex compact and is non-empty and bounded, then

(7)

Lemma 4.4 ([20]) Let, be normed spaces and be a convex cone in. A set-valued function satisfies the Jensen equation

(8)

if and only if there exists an additive set-valued function and a set such that

,.

We will extend the results of Aziz, Guerrero, Merentes and Sánchez given in [8] and [21] to set-valued functions of -bounded variation with respect to the weight function.

5. Main Results

Lemma 5.1 If such that satisfies the condition and

then is continuous.

Proof. Since, exists such that

(9)

for all partitions of, in particular given, we have

(10)

Since is convex -function, from the last inequality, we get

(11)

By (1),

(12)

This proves the continuity of at. Thus is continuous on.

Now, we are ready to formulate the main result of this work.

Main Theorem 5.2 Let, be normed spaces, be a convex cone in and be two convex -functions in, strictly increasing, that satisfy condition and such that there exists constants and with for all. If the Nemitskii operator generated by a set-valued function maps the space

into the space

and if it is globally Lipschitz, then the set-valued function satisfies the following conditions:

1) For every there exists, such that

(13)

2) There are functions and such that

(14)

Proof. 1) Since is globally Lipschitz, there exists a constant such that

(15)

Using the definitions of the operator and metric we have

where. In particular,

for all and, , where

.

Since and satisfy

(16)

we obtain

Therefore

(17)

Define the auxiliary function by:

(18)

Then and

Let us fix and define the functions by:

(19)

Then the functions and

(20)

From the definition of and, we have

(21)

From (16), we get

(22)

Hence,

(23)

Hence, substituting in inequality (5) the particular functions defined by (19) and taking in (23), we obtain

(24)

for all.

By Lemma 4.3 and the inequality (24), we have

for all.

Now, we have to consider the case. Define the function by

(25)

Then the function and

Let us fix and define the functions by

(26)

Then the functions (i = 1,2) and

Substituting and, and consider, we obtain

(27)

for all, where

By Lemma 4.3 and the above inequality, we get

for all. Define the function by

Hence

and, consequently, for every the function is continuous.

This completes the proof of part 1).

Now we shall prove that satisfies equality 2).

Let us fix such that. Since the Nemytskii operator is globally Lipschitzian, there exists a constant, such that

(28)

where. Define the function by

The function.

Let us fix and define the functions by

(29)

The functions and

.

Hence, substituting in the inequality (28) the particular functions defined by (29), we obtain

(30)

Since maps

intothen is continuous for all. Hence letting in the inequality (30), we get

(31)

for all and.

Thus for all, we have

(32)

Since is convex, we have

(33)

for all. Thus for all, the set-valued function satisfies the Jensen Equation (33). Now by Lemma 4.4, there exists an additive set-valued function and a set, such that

(34)

Substituting into inequality (13), we deduce that for all there exists, such that

consequently, for every the set-valued function is continuous, and

.

Since is additive and, then for all, thus.

The Nemytskii operator maps the space into the space, then

.

Consequently the set-valued function has to be of the form

where and

.

Theorem 5.3 Let, be normed spaces, a convex cone in and be two convex -functions in, strictly increassing, satisfying

condition and. If the Nemytskii operator generated by a set-valued function

maps the space

into the space and if it is globally Lipschizian, then the set-valued function satisfies the following condition

i.e., the Nemytskii operator is constant.

Proof. Since the Nemytskii operator is globally Lipschizian between and the space

, then there exists a constant, such that

(35)

Let us fix such that. Using the definitions of the operator and of the metric, we have

(36)

Define the auxiliary function by

The function and

Let us fix and define the functions by

(37)

The functions and

Hence, substituting in the inequality (36) the auxiliary functions defined by (37), we obtain

By Lemma 4.3 and the above inequality, we get

Since, letting in the above inequality, we have

Thus for all and for all, we get

Theorem 5.4 Let, be normed spaces, a convex cone in and be a convex - function in satisfying the condition. If the Nemytskii operator generated by a set-valued function maps the space into the space and if it is globally Lipschizian, then the left regularization of the function defined by

satisfies the following conditions:

• for all there exists, such that

, where is a linear continuous set-valued function, and.

Proof. We take, and define the auxiliary function by:

The function and

Let us fix and define the functions by

(38)

The functions and

(39)

From the definition of and, we obtain

(40)

Since the Nemytskii operator is globally Lipschitzian between

andthen there exists a constant, such that

for. By Lemma 4.3, substituting the particular functions defined by (38) in the above inequality, we obtain

(41)

for all. By Lemma 4.3, we get

(42)

for all and.

In the case where, by a similar reasoning as above, we obtain that there exists a constant, such that

(43)

Define the function by

(44)

Hence,

By passing to the limit in the inequality (41) by the inequality (43) and the definition of we have for all that there exists, such that

Now we shall prove that satisfies the following equality

where is a linear continuous set-valued functions, and

.

Let us fix such that. Define the partition of the interval by

The Nemytskii operator is globally Lipschitzian between and, then there exists a constant, such that

(45)

where

and

.

We define the function in the following way:

The function and

.

Let us fix and define the functions by:

(46)

The functions and

.

Substituting in the inequality (45) the particular functions defined in (46), we obtain

(47)

Since the Nemytskii operator maps the spaces into, then for all, the function. Letting in the inequality (47), we get

for all and. By passing to the limit when, we get

Since is a convex function, then

Thus for every, the set-valued function satisfies the Jensen equation. By Lemma 4.4 and by the property (a) previously established, we get that for all there exist an additive set-valued function and a set, such that

By the same reasoning as in the proof of Theorem 5.2, we obtain that

and.

6. Acknowledgements

This research was partly supported by CDCHTA of Universidad de Los Andes under the project NURR-C- 547-12-05-B.

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